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Divergence theorem states that the surface integral of a vector space field over a closed surface, known as the “flux” through the surface, is equal to the volume integral of the divergence and over region within the surface. It states intuitively that “the total of all sources of the field in a region (with sinks regarded as negative sources) gives the net flux out of the region.” One of the most important theorems in vector calculus is the “Divergence Theorem.” This theorem is applied to a wide range of difficult integral problems. In the divergence theorem, the surface integral is compared to the volume integral. The divergence theorem is a theorem in vector algebra that relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed.

Divergence Theorem is a theorem that compares the surface integral to the volume integral. It aids in determining the flux of a vector field through a closed area with the help of the volume encompassed by the vector field‘s divergence. In vector calculus, it is also known as Gauss’ Divergence Theorem.

Divergence Theorem Statement: Divergence theorem relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed by the surface. It says that the outward flux through such a closed surface is equal to the volume integral of the divergence of the field within the surface.

It is one of the most important theorems in calculus and is used to solve difficult integral problems. It is particularly useful in electrostatics and fluid dynamics.

The surface integral of a vector field A over a closed surface equals the volume integral of the divergence of a vector field A over the volume (V) enclosed by the closed surface, according to the Gauss Divergence Theorem.

\( \oint_{s}\vec{A}.d\vec{S} = \iint_{V}\int \left ( \vec{\bigtriangledown .\vec{A}} \right ) dV \)

Take a surface S that surrounds a volume V. Let vector A represent the vector field in the specified region. Let this volume be composed of many elementary volumes in the form of parallelepiped.

Consider the jth parallelopiped, which has a volume \( \bigtriangleup Vj \) and is bounded by a surface Sj with an area d vector Sj. The surface integral of vector A over surface Sj is denoted by

\( \oint_{s}\oint \vec{A}.d \vec{S_{j}} \)

Step 1: Consider the entire volume divided into elementary volumes I, II, and III, as shown in the figure above. The outward direction of elementary volume I is inward direction of elementary volume II, and the outward direction of elementary volume II is inward direction of elementary volume III, and so on.

Step 2: As a result, the sum of integrals of elementary volumes cancels out, and we are left with the surface integral arising from the surface of S.

\( \sum \oint_{Sj}\oint \vec{A}.d\vec{S_{j}} = \oint_{S}\oint \vec{A}.d\vec{S} \) …………….. (equation 1)

Step 3: When we multiply and divide the left-hand side of equation (1) by \( \bigtriangleup Vi \), we obtain

\( \oint_{S}\oint \vec{A}. d\vec{S} =\sum \frac{1}{\bigtriangleup V_{i}} \left ( \oint_{Si} \vec{A . d\vec{S}}\right )\bigtriangleup V_{i} \)

Step 4: Let us now suppose that the volume of surface S is divided into infinite elementary volumes, resulting in \( \bigtriangleup Vi =0 \)

\( \oint_{S}\oint \vec{A} .d \vec{S} =\lim_{\bigtriangleup V_{i \to 0}} \sum \frac{1}{\bigtriangleup V_{i}}\left ( \int_{S_{i}}\int \vec{A. d\vec{S_{i}}} \right )\bigtriangleup V_{i} \) …………(equation 2)

Step 5: Now,

\( \lim_{\bigtriangleup V_{i \to0 }} \left [ \frac{1}{\bigtriangleup V_{i}\left ( \oint_{Si}\vec{A .d\vec{S_{i}}} \right )} \right ] = \left ( \vec{V .\vec{A}} \right ) \)

Step 6: As a result, equation (2) becomes

\( \oint_{S}\oint \vec{A} . d \vec{S} =\sum \left ( \vec{\bigtriangledown . \vec{A}} \right ) \bigtriangleup v_{i} \)

Step 7: Given that \( \bigtriangleup Vi=0 \), as a result \( \sum Vi \) becomes integral over volume V.

\( \oint_{S}\oint \vec{A} .d \vec{S} =\iint_{V}\int \left ( \vec{\bigtriangledown . \vec{A}} \right ) d V \)

Let us learn about the Difference between Stokes Theorem and Divergence Theorem below:

Stokes theorem | Divergence Theorem |

The flux passing through a single surface is calculated by Stokes Theorem | The flux entering and exiting a solid through its surface is measured by the Divergence Theorem. |

According to Stoke’s Theorem, “the surface integral of a curl of a function across a surface limited by a closed surface is equal to the line integral of the specific vector function around that surface. | According to the divergence theorem, the total outward flow from a volume, as determined by the flux through its surface, may be calculated by adding up all the little amounts of outward flow in that volume using a triple integral of divergence. |

Stokes theorem relates surface integral to line integral. | Divergence theorem relates volume integral to surface integral. |

Stokes theorem formula:\(\mathop\oint\nolimits_C\overrightarrow{F\;}\;.\overrightarrow{dr}=\mathop\int\!\!\!\int \nolimits_S (\nabla \times \overrightarrow{F\;}\;).\overrightarrow {dS} \) | Divergence theorem formula:\( \oint_{s}\vec{A}.d\vec{S} =\iint_{V}\int\left(\vec{\bigtriangledown .\vec{A}} \right ) dV \) |

Divergence Theorem applications in calculus are given below:

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Example 1: Solve the, \( \iint_{s}F .dS \)

where \( F = \left ( 3x + z^{^{77}}, y^{2} – \sin x^{^{2}}z, xz + ye^{x^{5}}\right ) \) and

S is the box’s surface \( 0\leq x\leq 1, 0 \leq y\geq 3, 0\leq z\leq 2 \) Use the outward normal n

Solution: Given the ugliness of the vector field, computing this integral directly would be difficult. The divergence of F, on the other hand, is appealing:

div F = 3+2y +x

The divergence theorem is used to convert the surface integral to a triple integral.

\( \iint_{S}F . dS = \iint\int_{B} div F dV \)

Where B denotes the box

\( 0\leq x\leq 1, 0 \leq y\geq 3, 0\leq z\leq 2 \)

The triple integral of div F = 3+2y +x over the box B is computed:

\( \iint_{S}F .dS =\int_{0}^{1} \int_{0}^{3}\int_{0}^{2}\left ( 3+2y+x \right )dzdydx \)

\( \int_{0}^{1}\int_{0}^{3}\left ( 6+4y+2x \right )dy dx \)

\( \int_{0}^{1}\left ( 18+18+6x \right )dx \)

=36+3 =39

Example 2: Assume there are four stationary point charges in space, each of which has a charge of 0.002 Coulombs (C). The charges are (0,0,1), (1,1,4), (1,0,0), and (2,2,2). Let \( \vec{E} \) stand for the electrostatic field produced by these point charges. If S is a sphere of radius 2 oriented outward and centered at the origin then find:

\( \iint_{S}\vec{E} .d \vec{S} \)

Solution: Gauss’ law states that the flux of \( \vec{E} \) across S is equal to the total charge inside S divided by the electric constant. Because S has a radius of 2, only two charges are contained within it: the charge at (0, 1, 1) and the charge at (1, 0, 0). As a result, the total charge encompassed by S is 0.004 and, according to Gauss’ law,

\( \iint_{S}\vec{E} .d \vec{S} = \frac{0.004}{8.854\times 10^{-12}}\approx 4.418 V –m \)

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